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\newcommand{\Title}{Math 234, Midterm Exam, October 23, 2001}
\newif\ifanswer
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%\answerfalse

\def\Points{ Points}

\newcount\total
\def\Qi{30}     \advance\total by \Qi   %
\def\Qii{30}    \advance\total by \Qii  %
\def\Qiii{70}    \advance\total by \Qiii  %
\def\Qiv{30}     \advance\total by \Qiv   %
\def\Qv{30}    \advance\total by \Qv  %
\def\Qvi{30}   \advance\total by \Qvi %
\def\Qvii{30}  \advance\total by \Qvii %
%\def\Qviii{30}    \advance\total by \Qviii  %
%\def\Qix{15}     \advance\total by \Qx   %


\newcommand{\0}{\mathbf{0}}
\newcommand{\abf}{\mathbf{a}}
\newcommand{\cbf}{\mathbf{c}}
\newcommand{\hbf}{\mathbf{h}}
\newcommand{\ibf}{\mathbf{i}}
\newcommand{\jbf}{\mathbf{j}}
\newcommand{\kbf}{\mathbf{k}}
\newcommand{\rbf}{\mathbf{r}}
\newcommand{\ubf}{\mathbf{u}}
\newcommand{\vbf}{\mathbf{v}}
\newcommand{\Fbf}{\mathbf{F}}
\newcommand{\Nbf}{\mathbf{N}}
\newcommand{\Rbf}{\mathbf{R}}
\newcommand{\Tbf}{\mathbf{T}}
\newcommand{\p}{\partial}

\newcommand{\prob}[2]{\pagebreak[3]\noindent{\bf #1.} (#2 points.) }
\newcommand{\medskipnoindent}{\par\medskip\par\noindent}

\newcommand{\ds}{\displaystyle}

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\newcommand{\Answer}[1]{\par\medskip\par\noindent{\bf Answer: }{\small #1}\par\medskip\par}

\newcommand{\PageBreak}[1]{\par\bigskip\par\pagebreak[#1]\par}

\begin{document}

\vspace*{-0.5in}

\ifanswer
\else
Name \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
\fi
\medskip

\centerline{\large \Title}

\medskip

\ifanswer
 \centerline{\large Answers}
\medskip

\else



\medskip

\fi

\ifanswer\else
\begin{center}

{\em Circle your section.}

\begin{tabular}{|llr|llr|}
\hline
321  &  Cho &  7:45 T  &
322  &  Cho &  7:45 R  \\
323  &  Cho &  8:50 T  &
324  &  Cho &  8:50 R   \\
325  &  Raichev &  9:55 T  &
326  &  Raichev &  9:55 R  \\
329  &  Russell & 12:05 T  &
330  &  Russell & 12:05 R   \\
331  &  Russell &  1:20 T  &
332  &  Russell &  1:20 R    \\
333  &  Raichev &  2:25 T  &
334  &  Raichev &  2:25 R    \\ \hline
 \end{tabular}

\Large
\renewcommand{\arraystretch}{1.8}

\bigskip


\begin{tabular}{|l|l|@{\qquad\qquad}c|} \hline\hline
  I      & \Qi \Points    &     \\ \hline
  II     & \Qii \Points   &     \\ \hline
  III    & \Qiii \Points  &    \\ \hline
  IV     & \Qiv \Points   &     \\ \hline
  V      & \Qv \Points    &      \\ \hline
  VI     & \Qvi \Points   &      \\ \hline
  VII    & \Qvii \Points  &     \\ \hline
 % VIII   & \Qviii \Points  &     \\ \hline
 % IX    & \Qix \Points  &     \\ \hline
 % X     & \Qx \Points  &     \\ \hline
\hline \hline
Total & \the\total \Points &   \\ \hline \hline  % Total = 250
\end{tabular}
\end{center}
\bigskip

{\sc Show your reasoning. Your grade will be based in part
on the clarity of your writing. YOU MUST EXPLAIN HOW YOU GOT YOUR ANSWER.}


\newpage
\fi

\prob{I}{\Qi} (1) Find the gradient $\nabla U=\mathrm{grad}\, U$
of the function
$$
U=(x^2+y^2+z^2)^{-1/2}.
$$

\ifanswer \Answer{
$U_x=-(x^2+y^2+z^2)^{-3/2}x$,
$U_y=-(x^2+y^2+z^2)^{-3/2}y$, and
$U_z=-(x^2+y^2+z^2)^{-3/2}z$, so
$$
\nabla U=U_x\ibf+U_y\jbf+U_z\kbf =-\frac{x\ibf }{\rho^3}-\frac{y\jbf }{\rho^3}-\frac{z\kbf}{\rho^3}
$$
where
$\rho=(x^2+y^2+z^2)^{1/2}$.
}
\else
\vfill
\fi



\noindent(2) Find the total differential $dU$ of this function.


\ifanswer \Answer{
$\ds
d U=U_x\,dx+U_y\,dy+U_z\,dz =-\frac{x\,dx }{\rho^3}-\frac{y\,dy }{\rho^3}-\frac{z\,dz}{\rho^3}
$
}
\else

\vfill

\newpage
\fi


\prob{II}{\Qii}
A planet moves in space according to Newton's law
$
 m\, d^2\rbf/dt^2= - mGM|\rbf|^{-3}\rbf
$
where  $\rbf=x\ibf+y\jbf+z\kbf$ is the position vector
of the planet. Let $\vbf$ be the velocity vector,
$K=m|\vbf|^2/2$, and $W=-mGM/|\rbf|$.
Show that the quantity $E=K+W$  is constant.

\ifanswer \Answer{
By the previous problem
$$
\nabla W=-\nabla\left(\frac{mGM}{\rho}\right)
= \frac{mGM}{\rho^3}(x\ibf+ y\jbf + z\kbf)
=\frac{mGM}{|\rbf|^3}\rbf=-\Fbf=-m\abf.
$$
Now by the product rule for differentiation
$$
  \frac{dK}{dt} = \frac{m}{2}\frac{d}{dt}(\vbf\cdot\vbf) =
\frac{m}{2}\left(\frac{d\vbf}{dt}\cdot\vbf+\vbf\cdot\frac{d\vbf}{dt}\right)
=m\abf\cdot\vbf
$$
and by the chain rule
$$
  \frac{dW}{dt} = (\nabla W)\cdot\vbf= -m\abf\cdot\vbf
$$
so
$$
\frac{dE}{dt}=\frac{dK}{dt}+\frac{dW}{dt}=
m\abf\cdot\vbf-m\abf\cdot\vbf=0.
$$
}
\else
\newpage
\fi


\prob{III}{\Qiii} The position vector $\Rbf$ of a moving point is given by
$$
    \Rbf= (2t+3)\ibf+(t^2-1)\jbf.
$$
\ifanswer(This is example~3 on page 554 and problem~6 on page 564 of the text.)\fi


\medskipnoindent{\bf(1)} Find the velocity vector $\vbf$.

\ifanswer \Answer{
$\ds\vbf=\frac{d\Rbf}{dt}=2\ibf+2t\jbf$.
}
\else
\vfill
\fi


\medskipnoindent{\bf(2)} Find the acceleration vector $\abf$.

\ifanswer \Answer{
 $\ds\abf=\frac{d\vbf}{dt}=2\jbf$.
}
\else
\vfill
\fi

\medskipnoindent{\bf(3)} Find the speed.

\ifanswer \Answer{
$\ds\frac{ds}{dt}=|\vbf|=2\sqrt{1+t^2}$.
}
\else
\vfill
\fi


\medskipnoindent{\bf(4)} Find the unit tangent vector $\Tbf$.

\ifanswer \Answer{
$\ds\Tbf=\vbf/|\vbf|=\frac{\ibf}{\sqrt{1+t^2}}+\frac{t\jbf}{\sqrt{1+t^2}}$.
}
\else
\vfill

\newpage

\centerline{\small (Problem III continued)}

\medskip

\fi


\medskipnoindent{\bf(5)} Find the unit normal vector $\Nbf$.

\ifanswer \Answer{
The unit normal vector is perpendicular to the unit tangent vector
so
$$
  \Nbf= \pm\left(\frac{-t\ibf}{\sqrt{1+t^2}}+\frac{\jbf}{\sqrt{1+t^2}}\right)
$$
This vector points to the concave side of the curve so that $\ds\frac{d\Tbf}{ds}=\kappa\Nbf$
where $\kappa>0$. The curve is a parabola which opens up so the plus sign is correct.
}
\else
\vfill
\fi


\medskipnoindent{\bf(6)} Find the  normal and tangential components of the acceleration.

\ifanswer \Answer{
The tangential component is $\ds\abf\cdot\Tbf=2\jbf\cdot\Tbf=\frac{2t}{\sqrt{1+t^2}}$.
The normal component is $\ds\abf\cdot\Nbf=2\jbf\cdot\Nbf=\frac{2}{\sqrt{1+t^2}}$.
Hence $\ds\abf=\frac{2t}{\sqrt{1+t^2}}\Tbf+\frac{2}{\sqrt{1+t^2}}\Nbf$.
}
\else
\vfill
\fi

\medskipnoindent{\bf(7)} Find the  curvature.

\ifanswer \Answer{
 From the formula
$\ds\abf=\frac{d^2s}{dt^2}\Tbf+\kappa\left(\frac{ds}{dt}\right)^2\Nbf$
we get that
$$
  \kappa =\frac{\abf\cdot\Nbf}{(ds/dt)^2} =
\frac{2}{\sqrt{1+t^2}}\cdot\frac{1}{4(1+t^2)}=\frac{1}{2(1+t^2)^{3/2}}.
$$
}
\else

\vfill

\newpage
\fi



\prob{IV}{\Qiv} Find the local maxima and local minima of the function
$$
   f(x,y)= x^2-xy+y^2+2x+2y-4.
$$
\ifanswer (This is the example on page 611 of the text.)\fi

\ifanswer \Answer{$f_x=2x-y+2$, $f_y=-y+2y+2$, so $f_x=f_y=0$
if and only of $x=-2$, $y=-2$. For $(x,y)$ near  the point $(a,b)=(-2,-2)$
we have the quadratic approximation
$$
   Q(x,y)=-8+(x+2)^2-(x+2)(y+2)+(y+2)^2.
$$
(In this example $f(x,y)=Q(x,y)$.) The discriminant
$b^2-4ac=-3$ is negative so we have with a local maximum or a local minimum.
When $y=-2$ we have $Q(x,y)=-8+(x+2)^2\ge-8=Q(-2,-2)$ so $(-2,-2)$ is a local
minimum. (In fact it is a global minimum.) As $(-2,-2)$ is the only critical point
there is no other local extremum.
}
\else
\newpage
\fi


\prob{V}{\Qv} The temperature $T$ at any point $(x,y,z)$ in space is
$$
T=400xyz^2.
$$
Find the highest temperature on the unit sphere
$$
  x^2+y^2+z^2=1.
$$
\ifanswer (This is problem~57 on page 643 of the text.)\fi


\ifanswer \Answer{At the maximum we have $\nabla T= \lambda\nabla (x^2+y^2+z^2-1)$ or
\begin{eqnarray*}
  400 yz^2 &=& \lambda 2x \\
  400 xz^2 &=& \lambda 2y \\
  400 2xyz &=& \lambda 2z \\
   x^2+y^2+z^2 &=& 1.
\end{eqnarray*}
If $x=0$ or $y=0$ or $z=0$, then $T=0$. But the function $T$ takes
positive values so we may assume that $x\ne 0$ and $y\ne 0$ and $z\ne 0$.
Dividing the first two equations by the third gives
$$
   \frac{z}{2x}=\frac{x}{z},\qquad \frac{z}{2y}=\frac{y}{z}
$$
so $2x^2=z^2$ and $2y^2=z^2$. From the last equation we get
$1=x^2+y^2+z^2= 2z^2$ so $z=\pm1/\sqrt2$. The sign  of $x$ and $y$
must be the same (otherwise $T<0$) so the maximum occurs at one of the four points
$$
  (x,y,z)=\left(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt2}\right),\;
\left(-\frac{1}{2},-\frac{1}{2},\frac{1}{\sqrt2}\right),\;
\left(\frac{1}{2},\frac{1}{2},-\frac{1}{\sqrt2}\right),\;
\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{\sqrt2}\right).
$$
The value of $T$ at each of these four points is $400/8=50$.
}
\else
\newpage
\fi


\prob{VI}{\Qvi} {\bf(a)} Find an equation of the plane tangent to the surface
$$
    x^3+xy^2+y^3+z^3+1=0
$$
at the point $P_0(-2,1,2)$. \ifanswer{(This is problem~10 on page~641 of the text.)}\fi

\ifanswer \Answer{The point $P(x,y,z)$ lies on this plane if and only if
the vector from $P_0$ to $P$ is perpendicular to the gradient at $P_0$
of the defining function $F(x,y,z)=x^3+xy^2+y^3+z^3+1$. But
$$
  \nabla F=(3x^2+y^2)\ibf+(2xy+3y^2)\jbf+3z^2\kbf
$$
so
$$
\left.\nabla F\right|_{(-2,1,2)}=13\ibf-\jbf+12\kbf
$$
so the tangent plane is $13(x+2)-(y-1)+12(z-2)=0$.
}
\else

\vfill
\fi

\medskipnoindent{\bf(b)} Find equations of the straight line
through $P_0(-2,1,2)$ perpendicular
to the tangent plane found in part~(a).

\ifanswer \Answer{A point $P$ lies on the line if and only if the vector from
$P_0$ to $P$  is a multiple of $t\nabla F$ of the gradient vector
$\nabla F$ at $P_0$; hence the line has parametric equations
$x=-2+13t$, $y=1-t$, $z=2+12t$. Equations for the line
are
$$
 \frac{x+2}{13}=1-y=\frac{z-2}{12}.
$$
}
\else
\vfill

\newpage
\fi



\prob{VII}{\Qvii}{\bf(a)} Find the linear function which best approximates
$$
   f(x,y)= \frac{1}{1+x-y}
$$
near the point $(x,y)=(2,1)$.
\ifanswer(This is example~3 on page 589 of the text.)\fi

\ifanswer \Answer{
$\ds
L(x,y)=f(2,1)+f_x(2,1)(x-2)+f_y(2,1)(y-1)=\frac{1}{2}-\frac{x-2}{4}+\frac{y-1}{4}.
$
}
\else
\vfill
\fi

\medskipnoindent{\bf(b)} Find the tangent plane at the point $(x,y)=(2,1)$
to the surface
$$
z=\frac{1}{1+x-y}
$$



\ifanswer \Answer{$z=L(x,y)$, i.e. $\ds z=\frac{1}{2}-\frac{x-2}{4}+\frac{y-1}{4}$.
}
\else
\vfill
\fi


\ifanswer

\bigskip

\hrule

\begin{verbatim}

_________ Mon Oct 29 15:09:28 2001


--
There are  261 scores
grade      range       count      percent
   A     225...250     59         22.6%
  AB     215...224     16          6.1%
   B     200...214     30         11.5%
  BC     160...199     77         29.5%
   C     125...159     48         18.4%
   D     100...124     11          4.2%
   F       0... 99     20          7.7%
Mean score = 181.5. Mean grade = 2.61.
--
\end{verbatim}

\fi
\end{document}
